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minimum-increments-for-target-multiples-in-an-array | JAVA DP SOLUTION | java-dp-solution-by-subhankar752-83mr | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Subhankar752 | NORMAL | 2025-02-04T17:37:20.878062+00:00 | 2025-02-04T17:37:20.878062+00:00 | 31 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Dynamic Programming', 'Bitmask', 'Java'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Get the Intuition | Why DP Bitmask | 🏆 | get-the-intuition-why-dp-bitmask-by-sour-xtlg | Intuition- Why use Bitmask ??=> We have to select any comination of elemnts from target set, we can do that by seecting the set bit position elements of mask.
= | souraman19 | NORMAL | 2025-02-03T10:49:00.388666+00:00 | 2025-02-03T10:49:00.388666+00:00 | 106 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
## - Why use Bitmask ??
=> We have to select any comination of elemnts from target set, we can do that by seecting the set bit position elements of mask.
=> Again we can do OR opearion in case of numbers easily and can select larger subset ... | 1 | 0 | ['Array', 'Dynamic Programming', 'Bit Manipulation', 'Ordered Map', 'Ordered Set', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Simple Bitmask + DP | simple-bitmask-dp-by-22147407-hqwa | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | 22147407 | NORMAL | 2025-02-03T07:09:39.718378+00:00 | 2025-02-03T07:09:39.718378+00:00 | 40 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Dynamic Programming', 'Recursion', 'Memoization', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | ✅✅ C++ || Easy Explanation with intuition || Bitmask || DP ✅✅ | c-easy-explanation-with-intuition-bitmas-x5o9 | IntuitionThis problem is nothing more than dp with bitmasking, the idea comes because we dont know greedily that which number is it optimal to change nums[i] to | ssugamlm10 | NORMAL | 2025-02-03T06:55:57.512784+00:00 | 2025-02-03T06:55:57.512784+00:00 | 99 | false | # Intuition
This problem is nothing more than dp with bitmasking, the idea comes because we dont know greedily that which number is it optimal to change nums[i] to in order to make it a multiple of an element in target[j], or whether to leave it without making it a multiple since you have better options ahead!!! So, in... | 1 | 0 | ['Dynamic Programming', 'Greedy', 'Recursion', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ DP Solution | c-dp-solution-by-retire-pf0a | Code | Retire | NORMAL | 2025-02-03T01:26:51.932653+00:00 | 2025-02-03T01:26:51.932653+00:00 | 99 | false | # Code
```cpp []
typedef long long ll;
class Solution {
public:
int minimumIncrements(vector<int>& nums, vector<int>& target) {
int m = nums.size();
int n = target.size();
vector<ll> record(1 << n);
for (int i = 0; i < (1 << n); i++) {
ll need = 1;
for (int j ... | 1 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ || 100% || 40ms & 50.13mb || Optimized DP & Bitmasking for Minimum Increments to Cover Multiples | c-100-40ms-5013mb-optimized-dp-bitmaskin-1qzu | IntuitionWe observe that incrementing a number in nums allows it to "cover" one or more targets if, after increments, it becomes a multiple of those targets. In | 4fPYADGrHM | NORMAL | 2025-02-02T17:47:56.589029+00:00 | 2025-02-02T17:47:56.589029+00:00 | 34 | false | # Intuition
We observe that incrementing a number in `nums` allows it to "cover" one or more targets if, after increments, it becomes a multiple of those targets. In fact, if we take a subset of targets, we can require the number to be a multiple of the least common multiple (LCM) of that subset so it simultaneously co... | 1 | 0 | ['Math', 'Dynamic Programming', 'Greedy', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Easy Bitmask Solution | Python3 | Best Time Complexity | Beats 100% of Solutions | easy-bitmask-solution-python3-best-time-3nksc | ProblemThe problem requires us to ensure that each element in target has at least one multiple present in nums by incrementing elements of nums as needed. Our g | shreyvarshney1 | NORMAL | 2025-02-02T16:26:47.094143+00:00 | 2025-02-03T10:55:54.395962+00:00 | 87 | false | ## Problem
The problem requires us to ensure that each element in `target` has at least one multiple present in `nums` by incrementing elements of `nums` as needed. Our goal is to achieve this in the minimum number of operations.
### Key Observations:
1. If an element of `target` already has a multiple in `nums`, we d... | 1 | 0 | ['Dynamic Programming', 'Bitmask', 'Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP Bitmask Beat 100% Runtime and Memory | dp-bitmask-beat-100-runtime-and-memory-b-7thj | ApproachLet n be the size of nums, and t be the size of target. We use dynamic programming with two states dp[i][j], where 0≤i<n and 0≤j<2t, denoting the minimu | donbasta | NORMAL | 2025-02-02T14:36:29.190006+00:00 | 2025-02-02T14:36:29.190006+00:00 | 82 | false | # Approach
<!-- Describe your approach to solving the problem. -->
Let $n$ be the size of `nums`, and $t$ be the size of `target`. We use dynamic programming with two states $dp[i][j]$, where $0 \le i < n$ and $0 \le j < 2^t$, denoting the minimum number of increments so that each elements of `target` with indices in m... | 1 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Simple DP solution without LCM, O(1) space , beats 100%(Time & space) | dp-without-lcm-o1-space-by-saptak8837-cu3e | IntuitionTry to make each element in nums multiple of an element in targetApproachComplexity
Time complexity: O(n∗(2m))
Space complexity:O(1)
Code | saptak8837 | NORMAL | 2025-02-02T14:24:09.771121+00:00 | 2025-02-02T14:31:40.259064+00:00 | 97 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Try to make each element in nums multiple of an element in target
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: $$O(n*(2^m))$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- S... | 1 | 0 | ['Dynamic Programming', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Ugly Q3 but basically try all the possible choices and calculate LCM | ugly-q3-but-basically-try-all-the-possib-s2r9 | Code | nishsolvesalgo | NORMAL | 2025-02-02T04:05:34.742052+00:00 | 2025-02-02T04:05:34.742052+00:00 | 203 | false | # Code
```cpp []
class Solution {
public:
int minimumIncrements(vector<int>& nums, vector<int>& target) {
bool first = true, second = true, third = true, fourth = true;
for (int i = 0; i < target.size(); i++) {
if (i == 0) first = false;
if (i == 1) second = false;
... | 1 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Bitmask DP keep track of satisfied target and greedily choose LCM | bitmask-dp-keep-track-of-satisfied-targe-002v | null | theabbie | NORMAL | 2025-02-02T04:01:27.403095+00:00 | 2025-02-02T04:01:27.403095+00:00 | 330 | false | ```python3 []
import math
def gcd(a, b):
while b:
a, b = b, a % b
return a
def lcm(a, b):
return a * b // gcd(a, b)
class Solution:
def minimumIncrements(self, nums: List[int], target: List[int]) -> int:
k = len(target)
umask = (1 << k) - 1
n = len(nums)
... | 1 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ | DP + Bitmask | c-dp-bitmask-by-kena7-b9yb | Code | kenA7 | NORMAL | 2025-02-22T10:05:38.726086+00:00 | 2025-02-22T10:05:38.726086+00:00 | 9 | false |
# Code
```cpp []
class Solution {
public:
int dp[50001][16];
int minOps(int x, int mask, vector<int>&t)
{
long lcm=0;
for(int i=0;i<t.size();i++)
{
int bit=1&(mask>>i);
if(bit)
{
lcm=(lcm==0)?t[i]:(lcm*t[i])/__gcd(lcm,(long)t[i]);
... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ bit masks solution + lcm - DP | c-bit-masks-solution-lcm-dp-by-joelchaco-dh8u | IntuitionApproachComplexity
Time complexity:
O(N∗2N∗2M)
Space complexity:
O(N∗2M) can be improved to O(2M)
Code | joelchacon | NORMAL | 2025-02-15T18:58:52.388348+00:00 | 2025-02-15T18:58:52.388348+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
- $$O(N*2^N*2^M)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
- $$O(N*2^M)$$ can be improved to $$O(2^M)$$
... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Explained Clean Code | 27ms | DP + LCM + bitmask | explained-clean-code-27ms-dp-lcm-bitmask-7x4k | IntuitionUse Dynamic Programming with bitmasks to track which target numbers are covered by incrementing each number in nums, where "covered" means the incremen | ivangnilomedov | NORMAL | 2025-02-14T13:55:09.826820+00:00 | 2025-02-14T13:55:09.826820+00:00 | 7 | false | # Intuition
Use Dynamic Programming with bitmasks to track which target numbers are covered by incrementing each number in nums, where "covered" means the incremented number becomes a multiple of those target numbers.
# Approach
1. **Optimize Target Array**
- Remove duplicates.
2. **Precompute LCMs**
- Calculat... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP with bitmasking | dp-with-bitmasking-by-raikou_thunder-qdci | IntuitionApproachComplexity
Time complexity:O(256*n)
Space complexity:O(n*n)
Code | Raikou_Thunder | NORMAL | 2025-02-13T15:16:10.158336+00:00 | 2025-02-13T15:16:10.158336+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Appr... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Using 2D DP and BitMask + LCM AND GCD in C++ (100% faster) | using-2d-dp-and-bitmask-lcm-and-gcd-in-c-ax1x | IntuitionApproachComplexity
Time complexity:
O(n*16*16)
Space complexity:
O(n*16)
Code | Balram_54321 | NORMAL | 2025-02-09T09:49:17.775988+00:00 | 2025-02-09T09:49:17.775988+00:00 | 12 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
- O(n\*16*16)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
- O(n*16)
<!-- Add your space complexity here, e.g... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | (01 DP) + Bitmasking Without LCM approach | 01-dp-bitmasking-without-lcm-approach-by-vdrn | Code | spravinkumar9952 | NORMAL | 2025-02-08T12:33:48.642938+00:00 | 2025-02-08T12:33:48.642938+00:00 | 12 | false |
# Code
```cpp []
class Solution {
public:
int MX = 1e9;
int memo[50001][17];
int N;
int dfs(int ind, vector<int>& nums, vector<int>& target, int mask){
// Base case
if(ind >= nums.size()){
int targetMask = (1 << N)-1;
return mask == targetMask ? 0 : MX;
... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Simple DP+Bitmasking | simple-dpbitmasking-by-catchme999-yqae | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | catchme999 | NORMAL | 2025-02-08T11:56:57.905012+00:00 | 2025-02-08T11:56:57.905012+00:00 | 16 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Clean C++ solution O(N) | clean-c-solution-on-by-vishal_781-5umi | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | vishal_781 | NORMAL | 2025-02-08T08:42:21.949577+00:00 | 2025-02-08T08:42:21.949577+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | 5D dp.. | 5d-dp-by-joshuadlima-mhiw | Intuition0 -> 16 we get all possibilities for 4 numbers so i use that as bruteforcewe will keep track of whether each number in target has been satisfied or not | joshuadlima | NORMAL | 2025-02-07T14:56:22.535602+00:00 | 2025-02-07T14:56:22.535602+00:00 | 6 | false | # Intuition
0 -> 16 we get all possibilities for 4 numbers so i use that as bruteforce
we will keep track of whether each number in target has been satisfied or not
rest is basic math
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity her... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Dp + Bitmasking (Simplified) | dp-bitmasking-simplified-by-vineeth0904-8k87 | IntuitionApproachComplexity
Time complexity:
O(N * K * 2^k)
Space complexity:
Code | vineeth0904 | NORMAL | 2025-02-07T13:58:21.232846+00:00 | 2025-02-07T13:58:21.232846+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(N * K * 2^k)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
i... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ Dynamic Programming + Bit-Masking | c-dynamic-programming-bit-masking-by-sj4-pnwz | Intuitionkeep most things pre-computedCode | SJ4u | NORMAL | 2025-02-06T20:25:17.940085+00:00 | 2025-02-06T20:25:17.940085+00:00 | 10 | false | # Intuition
keep most things pre-computed
# Code
```cpp []
#define ll long long
class Solution {
public:
vector<vector<int>> pairOf1 = {{0}, {1}, {2}, {3}};
vector<vector<int>> pairOf2 = {{0, 1}, {0, 2}, {0, 3},
{1, 2}, {1, 3}, {2, 3}};
vector<vector<int>> pairOf3 = ... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Dp + bitmask | dp-bitmask-by-maxorgus-efer | Try to increase every number to the nearest multiplication of each of the traget and check whether we have had all the multiplactions.Upper limit of number of t | MaxOrgus | NORMAL | 2025-02-06T03:56:39.390723+00:00 | 2025-02-06T03:56:39.390723+00:00 | 10 | false | Try to increase every number to the nearest multiplication of each of the traget and check whether we have had all the multiplactions.
Upper limit of number of targets 4 does not look typically bitmask -- it kinda too small.
# Code
```python3 []
class Solution:
def minimumIncrements(self, nums: List[int], target:... | 0 | 0 | ['Dynamic Programming', 'Bitmask', 'Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | My Solution | my-solution-by-hope_ma-q72v | null | hope_ma | NORMAL | 2025-02-05T11:00:15.031384+00:00 | 2025-02-05T11:00:15.031384+00:00 | 12 | false | ```
/**
* Time Complexity: O(n * (3 ^ n_target))
* Space Complexity: O(2 ^ n_target)
* where `n` is the length of the vector `nums`
* `n_target` is the length of the vector `target`
*/
class Solution {
public:
int minimumIncrements(const vector<int> &nums, const vector<int> &target) {
constexpr int ra... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP Recursion Solution || O(n*4^k) || Easy to Understand | dp-recursion-solution-on4k-easy-to-under-zyvc | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | dnanper | NORMAL | 2025-02-05T08:37:29.273492+00:00 | 2025-02-05T08:37:29.273492+00:00 | 9 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Basic BitMask Application Solution | Beginner level Bit Mask | basic-bitmask-application-solution-begin-o8do | IntuitionBasic BitMask Dp Application
size of target is <<<< size of nums, generally a bit mask questionApproachwe will traverse on nums array and try to do ope | BOLTA7479 | NORMAL | 2025-02-05T04:17:18.249462+00:00 | 2025-02-05T04:22:44.173072+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Basic BitMask Dp Application
size of target is <<<< size of nums, generally a bit mask question
# Approach
<!-- Describe your approach to solving the problem. -->
we will traverse on nums array and try to do operations on it for a target wh... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | 😃🚀Classic Striver Take not Take O(n) (DP+BITMASKING)🚀😃 | classic-striver-take-not-take-on-dpbitma-pcf2 | Complexity
Time complexity:
O(2x∗x∗logM) + O(n∗2x∗2x)=O(n)
LCM_Precomputation + solve()
Space complexity:
O(n)
Code | Dynamic_Dishank | NORMAL | 2025-02-04T13:45:14.093239+00:00 | 2025-02-04T13:45:14.093239+00:00 | 24 | false | # Complexity
- Time complexity:
$$O(2^x * x * logM)$$ + $$O(n * 2^x * 2^x)$$=$$O(n)$$
LCM_Precomputation + solve()
- Space complexity:
$$O(n)$$
# Code
```cpp []
// Generalising the idea of take not_take
// Since some of the elements in the arr will not be used at all
// And some will serve some subset of target
// How... | 0 | 0 | ['Array', 'Dynamic Programming', 'Bit Manipulation', 'Number Theory', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Solution DP + Bitmasking +number theory | solution-dp-bitmasking-number-theory-by-2v4kv | IntuitionWe need to ensure that each element in target (b[]) has at least one multiple in nums (a[]). The key observation is that instead of making individual e | Happymood | NORMAL | 2025-02-04T09:42:13.624703+00:00 | 2025-02-04T09:42:13.624703+00:00 | 17 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We need to ensure that each element in target (b[]) has at least one multiple in nums (a[]). The key observation is that instead of making individual elements in a[] multiples of b[], we can use Least Common Multiple (LCM) to handle subsets... | 0 | 0 | ['Dynamic Programming', 'Bit Manipulation', 'Number Theory', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ [NON DP] approach that beats 99% | c-non-dp-approach-that-beats-99-by-thezi-3wte | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | thezi | NORMAL | 2025-02-03T19:46:06.770252+00:00 | 2025-02-03T19:46:06.770252+00:00 | 24 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Rust solution | rust-solution-by-abhineetraj1-scr7 | IntuitionThe problem asks for the minimum number of increments needed to make the elements of nums divisible by all the corresponding elements in target. The k | abhineetraj1 | NORMAL | 2025-02-03T18:02:26.619483+00:00 | 2025-02-03T18:02:26.619483+00:00 | 8 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem asks for the minimum number of increments needed to make the elements of nums divisible by all the corresponding elements in target. The key insight is that for a given number x and a set of targets, the minimum increments need... | 0 | 0 | ['Dynamic Programming', 'Greedy', 'Bit Manipulation', 'Rust'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP || bitmasking || C++. | dp-bitmasking-c-by-rishiinsane-f8s8 | IntuitionApproachComplexity
Time complexity:
o(n)
Space complexity:
O(16) ~ O(1)
Code | RishiINSANE | NORMAL | 2025-02-03T16:17:37.317550+00:00 | 2025-02-03T16:17:37.317550+00:00 | 19 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
o(n)
- Space complexity:
O(16) ~ O(1)
# Code
```cpp []
class Solution {
public:
int minimumIncrements(vector<int>& nums, vector<int>&... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Brute Force with Number Theory and Partitions | brute-force-with-number-theory-and-parti-adeo | ApproachConsider all the partitions and divide into groups based on the lcms of a certain partition set.
Consider all permutations of the lcm set and find out t | math_pi | NORMAL | 2025-02-03T16:12:43.596666+00:00 | 2025-02-03T16:12:43.596666+00:00 | 27 | false | # Approach
Consider all the partitions and divide into groups based on the lcms of a certain partition set.
Consider all permutations of the lcm set and find out the minimum answer via brute force.
# Complexity
- Time complexity:
$$O(n)$$
- Space complexity:
$$O(n)$$
# Code
```cpp []
class Solution {
public:
ve... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | [Python] concise combinations + bit-DP | python-concise-combinations-bit-dp-by-kn-p7ig | null | knatti | NORMAL | 2025-02-03T12:24:37.717140+00:00 | 2025-02-03T12:24:37.717140+00:00 | 27 | false | ```python3 []
class Solution:
def minimumIncrements(self, nums: List[int], target: List[int]) -> int:
N,T =len(nums), len(target)
multiples=defaultdict(list)
for mask in range(1,1<<T):
multiples[ lcm(*[n for i, n in enumerate(target) if (1<<i) &mask ])].append(mask)
@ca... | 0 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Simple Very Easy BIT MASK DP !! beats 100% | simple-very-easy-bit-mask-dp-beats-100-b-dopa | Complexity
Time complexity: O(1e7)
Space complexity: O(1e6)
Code | atulsoam5 | NORMAL | 2025-02-03T12:20:02.934689+00:00 | 2025-02-03T12:20:02.934689+00:00 | 14 | false | # Complexity
- Time complexity: O(1e7)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1e6)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int minimumIncrements(vector<int>& nums, vector<int>& target) {
int n = nums.size(), m = ... | 0 | 0 | ['Dynamic Programming', 'Bit Manipulation', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP Bitmasking soln | dp-bitmasking-soln-by-prikshit2803-i0e2 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Prikshit2803 | NORMAL | 2025-02-03T10:58:38.922189+00:00 | 2025-02-03T10:58:38.922189+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP+BitMask+ BinarySearch (Beats 100% in Time and Space) | dpbitmask-binarysearch-beats-100-in-time-hono | IntuitionDP+BitMask + BinarySearchApproachAdded in commentsComplexity
Time complexity: O(N*log(N))
Space complexity: O(N)
Code | ekalavya_pc | NORMAL | 2025-02-03T09:41:04.209820+00:00 | 2025-02-03T09:41:04.209820+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
DP+BitMask + BinarySearch
# Approach
<!-- Describe your approach to solving the problem. -->
Added in comments
# Complexity
- Time complexity: **O(N*log(N))**
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: **... | 0 | 0 | ['Binary Search', 'Dynamic Programming', 'Bitmask', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Sets + Binary Search + Generate Permutation | sets-binary-search-generate-permutation-j1pys | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kizu | NORMAL | 2025-02-03T07:54:40.374868+00:00 | 2025-02-03T07:54:40.374868+00:00 | 19 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Minimum Increments for Target Multiples in an Array ( Beats 100%) | minimum-increments-for-target-multiples-qly8h | Complexity
Time complexity: O(2^N)
Space complexity: O(N)
Code | Ansh1707 | NORMAL | 2025-02-02T17:52:13.717491+00:00 | 2025-02-02T17:52:13.717491+00:00 | 21 | false |
# Complexity
- Time complexity: O(2^N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python []
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
class Solution(object):
def minimumIncremen... | 0 | 0 | ['Math', 'Dynamic Programming', 'Bit Manipulation', 'Bitmask', 'Python'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ | Power set Solution | Easy | Self Understandable | c-power-set-solution-easy-self-understan-4p56 | IntuitionWe can get the minimum answer by taking lcm of none, one or more number of elements and then computing the minimun required to achieve that particluar | adithya_u_bhat | NORMAL | 2025-02-02T16:59:40.067828+00:00 | 2025-02-02T16:59:40.067828+00:00 | 27 | false | # Intuition
We can get the minimum answer by taking lcm of none, one or more number of elements and then computing the minimun required to achieve that particluar set of elements
# Approach
Try all combinations and along with all the priority combinations (the order in which they modify the nums is determined by prio... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | quick solution in java | quick-solution-in-java-by-klueqjnom3-7dho | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kLUeqJNoM3 | NORMAL | 2025-02-02T16:51:36.654776+00:00 | 2025-02-02T16:51:36.654776+00:00 | 37 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
minimum-increments-for-target-multiples-in-an-array | No DP/Bitmask | Bell Number | Store 4 lowest cost + Backtrack | bell-number-store-4-lowest-cost-backtrac-hevw | ApproachTargets may share or not share the same element in nums as min cost.
We group the targets such that only target in same group must share the same elemen | justinleung0204 | NORMAL | 2025-02-02T15:26:34.768027+00:00 | 2025-02-02T15:42:50.759492+00:00 | 40 | false | # Approach
<!-- Describe your approach to solving the problem. -->
Targets may share or not share the same element in nums as min cost.
We group the targets such that only target in same group must share the same element in nums, by considering the lcm of each group.
For the grouping, see [Bell Number](https://en.wiki... | 0 | 0 | ['Backtracking', 'Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Dynamic Programming | C++ | Solution with explanation | dynamic-programming-c-solution-with-expl-pifa | Solution
Also on my blog: https://www.zerotoexpert.blog/p/leetcode-weekly-contest-435
In this problem, we should find the minimum increments to meet the conditi | a_ck | NORMAL | 2025-02-02T13:00:47.525908+00:00 | 2025-02-02T13:00:47.525908+00:00 | 26 | false | # Solution
- Also on my blog: https://www.zerotoexpert.blog/p/leetcode-weekly-contest-435
In this problem, we should find the minimum increments to meet the condition. The critical point is that we only can increase the element. By this restriction, we can calculate the minimum value to make nums[i] to be the multiple... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Consider all cases(divide all, lcm all, 1/1/2, 1/rest, 2/2) w/ no index select duplications | consider-all-casesdivide-all-lcm-all-112-72oo | IntuitionGet minimum operations to achieve the goal, with given numbers (<= 4) and no index select duplications
Consider all cases(divide all, lcm all, 1/1/2, 1 | townizm | NORMAL | 2025-02-02T11:08:07.216313+00:00 | 2025-02-02T11:08:42.928855+00:00 | 30 | false | # Intuition
Get minimum operations to achieve the goal, with given numbers (<= 4) and no index select duplications
Consider all cases(divide all, lcm all, 1/1/2, 1/rest, 2/2)
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$... | 0 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | [Python] Top-Down DP + BitMask, Easy to understand with detailed explanation. | python-dp-bitmask-easy-to-understand-wit-y2b2 | IntuitionIn the contest, I notice that:
For each number in nums, we want to try if it can be a multiple to any value in the target. In this sense, we want to us | casterkiller | NORMAL | 2025-02-02T10:22:08.180222+00:00 | 2025-02-02T10:34:07.384703+00:00 | 38 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
In the contest, I notice that:
1. For each number in nums, we want to try if it can be a multiple to any value in the target. In this sense, we want to use DP, and a good way to mark the value in the target which have found a multiple in nu... | 0 | 0 | ['Dynamic Programming', 'Bit Manipulation', 'Bitmask', 'Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Beats 100% in Time and 100% in Space | beats-100-in-time-and-100-in-space-by-ha-qb47 | IntuitionAs we can see, bitmasking can be used so just brute force using bitmask DPComplexity
Time complexity:
Space complexity:
Code | harshkankhar1 | NORMAL | 2025-02-02T09:35:57.440390+00:00 | 2025-02-02T09:35:57.440390+00:00 | 41 | false | # Intuition
As we can see, bitmasking can be used so just brute force using bitmask DP
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int minimumIncrements... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | bitmask dp | bitmask-dp-by-ttn628826-00hg | Approach
LCM Calculation for Subsets:
We first precompute the LCM for every possible subset of the target array. A subset is represented by a bitmask, where e | ttn628826 | NORMAL | 2025-02-02T09:22:27.844992+00:00 | 2025-02-02T09:22:27.844992+00:00 | 23 | false | # Approach
1. LCM Calculation for Subsets:
- We first precompute the LCM for every possible subset of the target array. A subset is represented by a bitmask, where each bit indicates whether an element from the target set is included in the subset.
- The __computeSubsetLCMs__ function computes the LCM for eac... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | dp | dp-by-titu02-bkog | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Titu02 | NORMAL | 2025-02-02T09:16:57.398757+00:00 | 2025-02-02T09:16:57.398757+00:00 | 14 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Java Bitmask Dynamic Programming | java-bitmask-dynamic-programming-by-mot8-2izi | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | mot882000 | NORMAL | 2025-02-02T09:10:21.649030+00:00 | 2025-02-02T09:10:21.649030+00:00 | 46 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Dynamic Programming', 'Bitmask', 'Java'] | 0 |
minimum-increments-for-target-multiples-in-an-array | New Idea of Grouping elements | new-idea-of-grouping-elements-by-ishan-2-u4f4 | IntuitionSince size of target was small and greedy would not work was clearly visible, dp was an intutive approach.
Why greedy does not work ?
Suppose nums = [5 | Ishan-23 | NORMAL | 2025-02-02T08:26:31.601802+00:00 | 2025-02-02T08:26:31.601802+00:00 | 40 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Since size of target was small and greedy would not work was clearly visible, dp was an intutive approach.
Why greedy does not work ?
Suppose `nums = [5,7] and target = [4]`
Then Choosing `5` was a bad idea.
# Approach
<!-- Describe your ap... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | JAVA SOLUTION | java-solution-by-ashuramajestic-o60e | IntuitionThe problem requires us to modify the given numbers array so that each number can contribute to forming all values in the goal array using the Least Co | ashuramajestic | NORMAL | 2025-02-02T07:25:02.833777+00:00 | 2025-02-02T07:27:46.473261+00:00 | 51 | false | ## **Intuition**
The problem requires us to modify the given numbers array so that each number can contribute to forming all values in the goal array using the Least Common Multiple (LCM). Since modifying each number comes with a cost (the increment operation), our objective is to minimize this cost while ensuring th... | 0 | 0 | ['Bit Manipulation', 'Bitmask', 'Java'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Dynamic Programming + Python | dynamic-programming-python-by-leaf_light-9oda | IntuitionDP + BitmaskingApproachDP + BitmaskingComplexity
Time complexity:
O(n∗2t)
Space complexity:
O(n∗2t)
Code | leaf_light | NORMAL | 2025-02-02T07:01:37.614825+00:00 | 2025-02-02T07:01:37.614825+00:00 | 37 | false | # Intuition
DP + Bitmasking
# Approach
DP + Bitmasking
# Complexity
- Time complexity:
$$O(n*2^t)$$
- Space complexity:
$$O(n*2^t)$$
# Code
```python3 []
class Solution:
def minimumIncrements(self, nums: List[int], target: List[int]) -> int:
t = len(nums)
p = len(target)
lcm_dt = {... | 0 | 0 | ['Dynamic Programming', 'Bitmask', 'Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Dp - Bit Masking - Easy to Understand - Beats 100% | dp-bit-masking-easy-to-understand-beats-bo5ar | Code | UCJTBKG5qL | NORMAL | 2025-02-02T06:54:26.189782+00:00 | 2025-02-02T06:54:26.189782+00:00 | 28 | false |
# Code
```python3 []
class Solution:
def compute_lcm(self, a, b):
return (a*b)//math.gcd(a, b)
def minimumIncrements(self, nums: List[int], target: List[int]) -> int:
n = len(nums)
m = len(target)
if m == 0:
return 0
end = 1 << m
lcm = {}
for ... | 0 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP BRUTEFORCE || TOO EASY TO UNDERSTAND | dp-bruteforce-too-easy-to-understand-by-e5ouh | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kungfupandaisbackagain | NORMAL | 2025-02-02T06:45:38.788111+00:00 | 2025-02-02T06:46:07.302875+00:00 | 81 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Dynamic Programming', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP With Bitmasking | dp-with-bitmasking-by-arif2321-k8nl | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Arif2321 | NORMAL | 2025-02-02T05:55:59.490760+00:00 | 2025-02-02T05:55:59.490760+00:00 | 29 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Approach explained | Easy to understand | C++ | DP with Bitmask | easy-to-understand-c-dp-with-bitmask-by-jsyf3 | IntuitionSolve the DP for DP[i][mask]Approach
Each ith elment in nums can be taken or not taken to contribute to answer.
If it can be taken, it can contribute t | gamer37 | NORMAL | 2025-02-02T05:52:08.906726+00:00 | 2025-02-02T05:55:41.274684+00:00 | 51 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Solve the DP for DP[i][mask]
# Approach
<!-- Describe your approach to solving the problem. -->
- Each ith elment in nums can be taken or not taken to contribute to answer.
- If it can be taken, it can contribute to a subset of target. Th... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C# | c-by-7qlcxgljt2-t4am | Actually I didn't solve this question, but I see nobody post their C# answer. So I copy from Leetcode.cn. Answer is from here, I translate Java code to C# versi | 7qlcxGljt2 | NORMAL | 2025-02-02T05:43:30.025861+00:00 | 2025-02-02T05:43:30.025861+00:00 | 13 | false | Actually I didn't solve this question, but I see nobody post their C# answer. So I copy from Leetcode.cn. Answer is from [here](https://leetcode.cn/problems/minimum-increments-for-target-multiples-in-an-array/solutions/3061806/zi-ji-zhuang-ya-dpji-yi-hua-sou-suo-di-t-aeaj), I translate Java code to C# version.
# Compl... | 0 | 0 | ['C#'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Minimum Increments for Target Multiples in an Array | minimum-increments-for-target-multiples-0bdz8 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | subhampujari1 | NORMAL | 2025-02-02T04:54:14.820467+00:00 | 2025-02-02T04:54:14.820467+00:00 | 75 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP + Bitmask + Precomputation | Easy To Understand | Clean Code | dp-bitmask-precomputation-easy-to-unders-s6mu | Code | braindroid | NORMAL | 2025-02-02T04:41:55.369598+00:00 | 2025-02-02T04:41:55.369598+00:00 | 75 | false |
# Code
```cpp []
class Solution {
public:
vector<vector<long long>>sc;
vector<vector<long long>>dp;
int n;
int m;
long long solve(int i , int mask) {
if(i == n) {
if(mask == (1 << m) - 1) {
return 0;
}
return 1e9;
}
if(dp[... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | O(81 * n) bitmask dp solution | o81-n-bitmask-dp-solution-by-skuthuru-iis5 | IntuitionThere's only up to 4 target elements, this leads to an intuition of some bitmask dp solution.ApproachThe idea is to do a bitmask dp where we have dp[in | skuthuru | NORMAL | 2025-02-02T04:30:02.505860+00:00 | 2025-02-02T04:30:02.505860+00:00 | 58 | false | # Intuition
There's only up to 4 target elements, this leads to an intuition of some bitmask dp solution.
# Approach
The idea is to do a bitmask dp where we have dp[index][mask] storing the current mask of target elements we have taken the lcm of at this current index in nums. Next we will transition on all the submas... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | [JS] DFS over all reasonable solutions | js-dfs-over-all-reasonable-solutions-by-6q9d9 | IntuitionThe solution will involve adjusting up to 4 different numbers in nums. It will always be possible to come up with a solution by finding the LCM of ever | michaelm12358 | NORMAL | 2025-02-02T04:17:33.015177+00:00 | 2025-02-02T04:17:33.015177+00:00 | 74 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The solution will involve adjusting up to 4 different numbers in nums. It will always be possible to come up with **a** solution by finding the LCM of everything in target, but that may not be the **best** solution.
We can pessimistically ... | 0 | 0 | ['JavaScript'] | 0 |
minimum-increments-for-target-multiples-in-an-array | C++ double dfs | c-double-dfs-by-user5976fh-o90d | null | user5976fh | NORMAL | 2025-02-02T04:16:13.850790+00:00 | 2025-02-02T04:16:13.850790+00:00 | 79 | false | ```cpp []
class Solution {
public:
vector<int> v, t;
vector<vector<int>> grouped;
long long dfs(int i){
if (i == t.size()) return getMin();
long long ans = LLONG_MAX;
if (!grouped.empty()){
for (int y = 0; y < grouped.size(); ++y){
grouped[y].push_back(t[... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | LEETCODE WEEKLY 435 | Q3 | EASY TO UNDERSTAND | leetcode-weekly-435-q3-easy-to-understan-8l02 | IntuitionThe problem requires modifying elements in nums such that their least common multiple (LCM) covers all elements in target.
We need to increment numbers | prybruhta | NORMAL | 2025-02-02T04:10:26.430469+00:00 | 2025-02-02T04:10:26.430469+00:00 | 139 | false | # **Intuition**
The problem requires modifying elements in `nums` such that their **least common multiple (LCM)** covers all elements in `target`.
- We need to **increment numbers minimally** so that at least one subset of `nums` has an LCM equal to some subset of `target`.
- The approach revolves around **bitmaski... | 0 | 0 | ['Dynamic Programming', 'Bit Manipulation', 'C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Python Hard | python-hard-by-lucasschnee-09wi | null | lucasschnee | NORMAL | 2025-02-02T04:09:56.328334+00:00 | 2025-02-02T04:09:56.328334+00:00 | 79 | false | ```python3 []
class Solution:
def minimumIncrements(self, nums: List[int], target: List[int]) -> int:
'''
target.length <= 4
'''
nums.sort()
target.sort()
N = len(nums)
M = len(target)
INF = 10 ** 20
def get_min_multi... | 0 | 0 | ['Python3'] | 0 |
minimum-increments-for-target-multiples-in-an-array | Group by LCM | group-by-lcm-by-jianstanleya-grg6 | IntuitionNotice how we can "match" each number in target individually or multiple at once using their LCM.ApproachWe iterate over all unique partitions of targe | jianstanleya | NORMAL | 2025-02-02T04:05:57.759442+00:00 | 2025-02-02T04:06:43.881706+00:00 | 95 | false | # Intuition
Notice how we can "match" each number in `target` individually or multiple at once using their LCM.
# Approach
We iterate over all unique partitions of `target`. For each subset in each partition, we find the best number to match with in `nums`. Each number should only be used once!
# Complexity
- Time co... | 0 | 0 | ['C++'] | 0 |
minimum-increments-for-target-multiples-in-an-array | DP for Bitmasks | dp-for-bitmasks-by-whoawhoawhoa-gmz7 | IntuitionGiven that target array length is 4 at maximum we can employ bitmasking here.ApproachFirst step - we find lcm map containing lcm for each combination o | whoawhoawhoa | NORMAL | 2025-02-02T04:01:03.480000+00:00 | 2025-02-02T04:07:25.083615+00:00 | 164 | false | # Intuition
Given that target array length is 4 at maximum we can employ bitmasking here.
# Approach
First step - we find lcm map containing lcm for each combination of numbers in target array.
Then we do Top-Down DP keeping the state of mask / index.
At each index our process is:
1. Try to keep `nums[ix]` to be the s... | 0 | 0 | ['Math', 'Bit Manipulation', 'Bitmask', 'Java'] | 1 |
special-positions-in-a-binary-matrix | C++ 2 passes | c-2-passes-by-votrubac-4vhx | First pass, count number of ones in rows and cols. \n\nSecond pass, go through the matrix, and for each one (mat[i][j] == 1) check it it\'s alone in the corresp | votrubac | NORMAL | 2020-09-13T04:02:31.690736+00:00 | 2020-09-14T18:10:33.178652+00:00 | 10,539 | false | First pass, count number of ones in `rows` and `cols`. \n\nSecond pass, go through the matrix, and for each one (`mat[i][j] == 1`) check it it\'s alone in the corresponding row (`rows[i] == 1`) and column (`cols[j] == 1`).\n\nThis is a copycat question to [531. Lonely Pixel I](https://leetcode.com/problems/lonely-pixel... | 146 | 3 | [] | 17 |
special-positions-in-a-binary-matrix | ✅Beats 100% - Explained with [ Video ] - Check Row and Column - C++/Java/Python/JS - Visualized | beats-100-explained-with-video-check-row-0jq6 | \n\n# YouTube Video Explanation:\n\nhttps://youtu.be/GI7EWjkXLnY\n **If you want a video for this question please write in the comments** \n\n\uD83D\uDD25 Pleas | lancertech6 | NORMAL | 2023-12-13T02:14:41.735819+00:00 | 2023-12-13T02:49:23.658357+00:00 | 19,332 | false | \n\n# YouTube Video Explanation:\n\n[https://youtu.be/GI7EWjkXLnY](https://youtu.be/GI7EWjkXLnY)\n<!-- **If you want a video for this question please write in the comments** -->\n\n**... | 87 | 2 | ['Array', 'Matrix', 'Python', 'C++', 'Java', 'JavaScript'] | 9 |
special-positions-in-a-binary-matrix | Java Simple Loop | java-simple-loop-by-hobiter-u97k | \n public int numSpecial(int[][] mat) {\n int m = mat.length, n = mat[0].length, res = 0, col[] = new int[n], row[] = new int[m];\n for (int i | hobiter | NORMAL | 2020-09-13T04:02:48.095117+00:00 | 2020-09-15T05:04:13.349551+00:00 | 5,651 | false | ```\n public int numSpecial(int[][] mat) {\n int m = mat.length, n = mat[0].length, res = 0, col[] = new int[n], row[] = new int[m];\n for (int i = 0; i < m; i++) \n for (int j = 0; j < n; j++) \n if (mat[i][j] == 1){\n col[j]++;\n row[i]+... | 85 | 3 | [] | 9 |
special-positions-in-a-binary-matrix | 【Video】Give me 5 minutes - How we think about a solution | video-give-me-5-minutes-how-we-think-abo-hrkz | Intuition\nCheck row and column direction.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/qTAh4HGfUHk\n\n\u25A0 Timeline of the video\n\n0:05 Explain algorithm | niits | NORMAL | 2023-12-13T01:33:13.578775+00:00 | 2023-12-14T04:13:27.411387+00:00 | 8,080 | false | # Intuition\nCheck row and column direction.\n\n---\n\n# Solution Video\n\nhttps://youtu.be/qTAh4HGfUHk\n\n\u25A0 Timeline of the video\n\n`0:05` Explain algorithm of the first step\n`1:06` Explain algorithm of the second step\n`2:36` Coding\n`4:17` Time Complexity and Space Complexity\n`4:55` Step by step algorithm of... | 62 | 0 | ['C++', 'Java', 'Python3', 'JavaScript'] | 7 |
special-positions-in-a-binary-matrix | ✅☑[C++/Java/Python/JavaScript] || 2 Approaches || EXPLAINED🔥 | cjavapythonjavascript-2-approaches-expla-73ag | PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n(Also explained in the code)\n\n#### Approach 1(Brute Force)\n1. It iterates through each element of the ma | MarkSPhilip31 | NORMAL | 2023-12-13T00:32:23.798285+00:00 | 2023-12-13T03:05:56.978491+00:00 | 9,028 | false | # PLEASE UPVOTE IF IT HELPED\n\n---\n\n\n# Approaches\n**(Also explained in the code)**\n\n#### ***Approach 1(Brute Force)***\n1. It iterates through each element of the matrix, checking if it equals 1.\n1. For each element that equals 1, it checks the entire row and column to see if there are any other \'1\'s apart fr... | 48 | 3 | ['Array', 'C', 'Matrix', 'C++', 'Java', 'Python3', 'JavaScript'] | 6 |
special-positions-in-a-binary-matrix | [Python] Easy to Understand Solution | python-easy-to-understand-solution-by-ch-v4yj | \nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n n = len(mat)\n m = len(mat[0])\n #taking transpose of matrix \n | chandreshwar | NORMAL | 2020-09-13T04:02:09.314535+00:00 | 2020-09-13T04:02:09.314578+00:00 | 3,649 | false | ```\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n n = len(mat)\n m = len(mat[0])\n #taking transpose of matrix \n lst = [[mat[j][i] for j in range(n)] for i in range(m)] \n res = 0\n for i in range(n):\n for j in range(m):\n ... | 42 | 6 | [] | 7 |
special-positions-in-a-binary-matrix | Python easy to understand 100% faster | python-easy-to-understand-100-faster-by-7rjfw | \n\ndef column(matrix, i):\n return [row[i] for row in matrix]\nclass Solution:\n \n def numSpecial(self, mat: List[List[int]]) -> int:\n c= | jatindscl | NORMAL | 2020-09-13T04:13:34.761215+00:00 | 2020-09-13T04:13:34.761300+00:00 | 3,271 | false | \n```\ndef column(matrix, i):\n return [row[i] for row in matrix]\nclass Solution:\n \n def numSpecial(self, mat: List[List[int]]) -> int:\n c=0\n for i in range(len(mat)):\n if(mat[i].count(1)==1):\n i=mat[i].index(1)\n col=column(mat,i)\n ... | 27 | 1 | ['Python', 'Python3'] | 5 |
special-positions-in-a-binary-matrix | 💯Faster✅💯 Lesser✅2 Methods🔥Brute Force🔥Hashing🔥Python🐍Java☕C++✅C📈 | faster-lesser2-methodsbrute-forcehashing-ibmj | \uD83D\uDE80 Hi, I\'m Mohammed Raziullah Ansari, and I\'m excited to share 2 ways to solve this question with detailed explanation of each approach:\n\n# Proble | Mohammed_Raziullah_Ansari | NORMAL | 2023-12-13T04:37:18.379851+00:00 | 2023-12-13T09:22:32.078007+00:00 | 2,793 | false | # \uD83D\uDE80 Hi, I\'m [Mohammed Raziullah Ansari](https://leetcode.com/Mohammed_Raziullah_Ansari/), and I\'m excited to share 2 ways to solve this question with detailed explanation of each approach:\n\n# Problem Explaination: \nThe problem involves finding the number of special positions in a binary matrix. A specia... | 24 | 0 | ['Array', 'Hash Table', 'C', 'Matrix', 'C++', 'Java', 'Python3'] | 4 |
special-positions-in-a-binary-matrix | [C++] SImple Approach Explained and Code and Complexity | c-simple-approach-explained-and-code-and-qmyy | This problem is very easy to think. Please do not go to code directly, take steps as hint and try to implement it yourself :)\nSteps:\n\t1. Get the sum of all t | rahulanandyadav2000 | NORMAL | 2020-09-13T04:17:16.649477+00:00 | 2020-09-13T04:17:16.649518+00:00 | 2,057 | false | This problem is very easy to think. Please do not go to code directly, take steps as hint and try to implement it yourself :)\nSteps:\n\t1. Get the sum of all the rows\n\t2. Get the sum of all the coloumns\n\t3. Traverse the matrix and whenever encounter 1 check its row and col sum, if 1 then increment count by 1.\n\t4... | 18 | 0 | ['C'] | 4 |
special-positions-in-a-binary-matrix | 🚀🚀🚀EASY SOLUTiON EVER !!! || Java || C++ || Python || JS || C# || Ruby || Go 🚀🚀🚀 by PRODONiK | easy-solution-ever-java-c-python-js-c-ru-apwg | Intuition\nThe problem aims to find the number of special positions in a given matrix. A position is considered special if the element at that position is 1, an | prodonik | NORMAL | 2023-12-13T00:19:20.588235+00:00 | 2023-12-13T01:38:46.842848+00:00 | 2,893 | false | # Intuition\nThe problem aims to find the number of special positions in a given matrix. A position is considered special if the element at that position is 1, and there is no other 1 in the same row and the same column.\n\n# Approach\nThe `numSpecial` method iterates over each row in the matrix. For each element with ... | 17 | 0 | ['C++', 'Java', 'Go', 'Python3', 'Ruby', 'JavaScript', 'C#'] | 4 |
special-positions-in-a-binary-matrix | Python | Readable Solution with Comments | python-readable-solution-with-comments-b-kshe | \nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n \n res = 0\n rows = len(mat)\n cols = len(mat[0])\n | jgroszew | NORMAL | 2022-01-08T19:18:36.048619+00:00 | 2022-01-08T19:18:36.048663+00:00 | 819 | false | ```\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n \n res = 0\n rows = len(mat)\n cols = len(mat[0])\n \n rowtotals = []\n coltotals = []\n \n # Store the sum of each row\n for row in mat:\n rowtotals.append(sum... | 13 | 0 | ['Python'] | 1 |
special-positions-in-a-binary-matrix | 🚀✅🚀 100% beats Users by TC || Java, C#, Python, Rust, C++, C, Ruby, Go|| 🔥🔥 | 100-beats-users-by-tc-java-c-python-rust-4e29 | Intuition\nThe problem seems to be asking for the number of special positions in a binary matrix. A position is considered special if it contains a 1 and the su | ilxamovic | NORMAL | 2023-12-13T04:18:42.500306+00:00 | 2023-12-13T04:19:32.614821+00:00 | 1,053 | false | # Intuition\nThe problem seems to be asking for the number of special positions in a binary matrix. A position is considered special if it contains a 1 and the sum of elements in its row and column is 1.\n\n# Approach\nThe provided code iterates through each element of the matrix and checks if it is 1. If it is, it cal... | 10 | 0 | ['C', 'C++', 'Java', 'Go', 'Python3', 'Rust', 'Ruby', 'C#'] | 1 |
special-positions-in-a-binary-matrix | C++/Python one pass||0 ms Beats 100% | cpython-one-pass0-ms-beats-100-by-anwend-uz5s | Intuition\n Describe your first thoughts on how to solve this problem. \nCheck which rows & columns have exactly one 1!\n\nThe revised C++ code simplifies the l | anwendeng | NORMAL | 2023-12-13T02:12:36.685440+00:00 | 2023-12-14T22:43:08.859049+00:00 | 3,000 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nCheck which rows & columns have exactly one 1!\n\nThe revised C++ code simplifies the loop. The unnecessary copying for column vector is omitted which becomes a fast code, runs in 0 ms and beats 100%!\n\nPython code is in the same manner ... | 10 | 0 | ['C', 'Matrix', 'Python3'] | 2 |
special-positions-in-a-binary-matrix | Java | Simple | Clean code | java-simple-clean-code-by-judgementdey-vts7 | Complexity\n- Time complexity: O(m*n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(m+n)\n Add your space complexity here, e.g. O(n) \n\n | judgementdey | NORMAL | 2023-12-13T01:42:04.886204+00:00 | 2023-12-13T01:42:04.886233+00:00 | 855 | false | # Complexity\n- Time complexity: $$O(m*n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(m+n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int numSpecial(int[][] mat) {\n var m = mat.length;\n var n = mat[0].length;\n\n ... | 9 | 0 | ['Array', 'Matrix', 'Java'] | 1 |
special-positions-in-a-binary-matrix | 🚀🚀 Beats 95% | Easy To Understand | Fully Explained 😎💖 | beats-95-easy-to-understand-fully-explai-wc0a | Intuition\n- We want to find special positions in a binary matrix.\n- A position is special if the element at that position is 1 and all other elements in its r | The_Eternal_Soul | NORMAL | 2023-12-13T02:34:32.139050+00:00 | 2023-12-13T02:34:32.139076+00:00 | 956 | false | # Intuition\n- We want to find special positions in a binary matrix.\n- A position is special if the element at that position is 1 and all other elements in its row and column are 0.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Iterate through the matrix to find the positions w... | 8 | 0 | ['Array', 'C', 'Matrix', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 1 |
special-positions-in-a-binary-matrix | [Python 3] One pass || 155ms || beats 99% | python-3-one-pass-155ms-beats-99-by-your-4qqw | python3 []\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n res = 0\n rows = [[] for _ in range(len(mat))]\n cols | yourick | NORMAL | 2023-06-28T13:08:50.668536+00:00 | 2023-12-13T00:05:59.678439+00:00 | 784 | false | ```python3 []\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n res = 0\n rows = [[] for _ in range(len(mat))]\n cols = [0] * len(mat[0])\n \n for i in range(len(mat)):\n for j in range(len(mat[0])):\n if mat[i][j]:\n ... | 8 | 0 | ['Matrix', 'Python', 'Python3'] | 0 |
special-positions-in-a-binary-matrix | C++ || Matrix || Short and Simple | c-matrix-short-and-simple-by-user3405ye-mcr2 | Please upvote if you find the solution clean and concise.\nThank You!!\n\n\nclass Solution {\npublic:\n int numSpecial(vector<vector<int>>& mat) {\n i | user3405Ye | NORMAL | 2023-04-24T05:01:35.033878+00:00 | 2023-04-24T05:01:35.033926+00:00 | 746 | false | Please upvote if you find the solution clean and concise.\n***Thank You!!***\n\n```\nclass Solution {\npublic:\n int numSpecial(vector<vector<int>>& mat) {\n int m = mat.size(), n = mat[0].size(), res = 0;\n vector<int> rows(m), cols(n);\n for(int i=0;i<m;i++) {\n for(int j=0;j<n;j++)... | 8 | 0 | ['C', 'Matrix', 'C++'] | 0 |
special-positions-in-a-binary-matrix | a few solutions | a-few-solutions-by-claytonjwong-udea | Accumulate the sum of values for each ith row and jth column. Traverse each cell i, j of the input A and increment the count cnt if and only if A[i][j], row[i] | claytonjwong | NORMAL | 2020-09-13T04:00:31.779041+00:00 | 2023-01-01T23:54:41.859548+00:00 | 763 | false | Accumulate the sum of values for each `i`<sup>th</sup> row and `j`<sup>th</sup> column. Traverse each cell `i`, `j` of the input `A` and increment the count `cnt` if and only if `A[i][j]`, `row[i]`, and `col[j]` are all equal-to 1.\n\n---\n\n*Kotlin*\n```\nclass Solution {\n fun numSpecial(A: Array<IntArray>): Int ... | 8 | 1 | [] | 1 |
special-positions-in-a-binary-matrix | Go one pass 16 ms, 6.1 MB | go-one-pass-16-ms-61-mb-by-vashik-e2m4 | \nfunc numSpecial(mat [][]int) int {\n rows, columns := len(mat),len(mat[0])\n specials := 0\n for i := 0; i < rows; i++ {\n checkCol := -1\n | vashik | NORMAL | 2020-09-14T12:15:24.141329+00:00 | 2020-09-14T12:26:45.004033+00:00 | 401 | false | ```\nfunc numSpecial(mat [][]int) int {\n rows, columns := len(mat),len(mat[0])\n specials := 0\n for i := 0; i < rows; i++ {\n checkCol := -1\n for j := 0; j < columns; j++ {\n if mat[i][j] == 1 {\n if checkCol == -1 {\n checkCol = j\n ... | 7 | 0 | ['Go'] | 1 |
special-positions-in-a-binary-matrix | ⭐✅|| STORING EACH ROW AND COL SUM || WELL EXPLAINED (HINDI) || BEATS 100% || ✅⭐ | storing-each-row-and-col-sum-well-explai-nz5m | \n Describe your first thoughts on how to solve this problem. \n\n# ALGORITHM\n Describe your approach to solving the problem. \n1)Matrix ke dimensions nikalo:\ | Sautramani | NORMAL | 2023-12-13T04:18:22.651848+00:00 | 2023-12-13T04:30:02.380696+00:00 | 794 | false | \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# ALGORITHM\n<!-- Describe your approach to solving the problem. -->\n1)**Matrix ke dimensions nikalo**:\nInput matrix mat ki rows (n) aur columns (m) ka pata lagayein.\n\n2)**Variables ko initialize karein**:\nans naamak ek variable ko 0 se initi... | 6 | 0 | ['Array', 'Matrix', 'Python', 'C++', 'Java', 'Python3'] | 3 |
special-positions-in-a-binary-matrix | EASY DECEMBER ⛄|| BEGINEER FRIENDLY APPROACH 🔥 || C++ | easy-december-begineer-friendly-approach-wmep | Intuition and Appraoch :\nSeems like it all easy december going.. \u26C4\n\nSo, this problem we will do matrix traversal in both row wise and column wise. now | DEvilBackInGame | NORMAL | 2023-12-13T03:46:58.366191+00:00 | 2023-12-13T14:04:35.013169+00:00 | 586 | false | # Intuition and Appraoch :\nSeems like it all easy december going.. \u26C4\n\nSo, this problem we will do matrix traversal in both row wise and column wise. now how we doing lets discuss .\ni will discuss wrt array so that its easy to understand.\n1. First i am assigning` length of array` or can say `number of rows` w... | 6 | 0 | ['Array', 'Matrix', 'C++'] | 1 |
special-positions-in-a-binary-matrix | Simple beginner friendly solution involving unordered maps | simple-beginner-friendly-solution-involv-puhb | Intuition\n Describe your first thoughts on how to solve this problem. \nAs question demands checking if the whole row and column of the current element has a 1 | maxheap | NORMAL | 2023-12-13T04:08:42.451092+00:00 | 2023-12-13T04:08:42.451141+00:00 | 1,045 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAs question demands checking if the whole row and column of the current element has a 1 anywhere , we would obviously need to precompute all the visited elements(tbp 1 here).\n\nElements visited here are tracked via two seperate unordered... | 5 | 0 | ['C++'] | 0 |
special-positions-in-a-binary-matrix | C# Solution for Special Positions In A Binary Matrix Problem | c-solution-for-special-positions-in-a-bi-vfyy | Intuition\n Describe your first thoughts on how to solve this problem. \n1.\tIterate through the matrix and count the number of ones in each row and each column | Aman_Raj_Sinha | NORMAL | 2023-12-13T03:04:20.992228+00:00 | 2023-12-13T03:04:20.992263+00:00 | 281 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.\tIterate through the matrix and count the number of ones in each row and each column.\n2.\tCheck each element of the matrix and verify if it is a special position according to the conditions specified.\n\n# Approach\n<!-- Describe your... | 5 | 0 | ['C#'] | 0 |
special-positions-in-a-binary-matrix | [C++] Two Simple Solutions | c-two-simple-solutions-by-pankajgupta20-pmu6 | Solution: 1\nTime Complexity: O(n^3)\nSpace Complexity: O(1)\n\n\tclass Solution {\n\tpublic:\n\t\tint numSpecial(vector>& mat) {\n\t\t\tint nRows = mat.size(); | pankajgupta20 | NORMAL | 2021-05-14T19:16:05.532687+00:00 | 2021-05-14T19:16:05.532735+00:00 | 941 | false | ##### Solution: 1\nTime Complexity: O(n^3)\nSpace Complexity: O(1)\n\n\tclass Solution {\n\tpublic:\n\t\tint numSpecial(vector<vector<int>>& mat) {\n\t\t\tint nRows = mat.size();\n\t\t\tint nCols = mat[0].size();\n\t\t\tint res = 0;\n\t\t\tfor(int i = 0; i < nRows; i++){\n\t\t\t\tfor(int j = 0; j < nCols; j++){\n\t\t\t... | 5 | 0 | ['C', 'C++'] | 0 |
special-positions-in-a-binary-matrix | Simple (5 line) efficient (faster than 99.8%) Python solution | simple-5-line-efficient-faster-than-998-6rr98 | \nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n count = 0\n cols = list(zip(*mat))\n for row in mat:\n | Kartheyan | NORMAL | 2020-10-13T12:17:44.142669+00:00 | 2020-10-13T12:17:44.142700+00:00 | 342 | false | ```\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n count = 0\n cols = list(zip(*mat))\n for row in mat:\n if sum(row)==1 and sum(cols[row.index(1)])==1:\n count+=1\n return count\n``` | 5 | 0 | [] | 0 |
special-positions-in-a-binary-matrix | JavaScript ~[76 ms, 40.2 MB] | javascript-76-ms-402-mb-by-dl90-iiue | \uD83D\uDE80\njs\nvar numSpecial = function (mat) {\n const rows = mat.length\n const cols = mat[0].length\n const rowArr = new Int8Array(rows)\n const colA | dl90 | NORMAL | 2020-09-16T06:37:08.997599+00:00 | 2020-09-16T06:37:56.966615+00:00 | 458 | false | \uD83D\uDE80\n```js\nvar numSpecial = function (mat) {\n const rows = mat.length\n const cols = mat[0].length\n const rowArr = new Int8Array(rows)\n const colArr = new Int8Array(cols)\n const set = new Set()\n\n for (let i = 0; i < rows; i++) {\n for (let j = 0; j < cols; j++) {\n if (mat[i][j]) {\n ... | 5 | 0 | ['JavaScript'] | 0 |
special-positions-in-a-binary-matrix | Java | O (M*N) Time | O (M+N) space | java-o-mn-time-o-mn-space-by-arizonazerv-zzyk | Intuition was to do exactly as the problem asked, count the number of 1\'s in every row and column, store them in a separate array and later whenever we encount | arizonazervas | NORMAL | 2020-09-13T04:10:43.427235+00:00 | 2020-09-13T04:10:43.427277+00:00 | 577 | false | * Intuition was to do exactly as the problem asked, count the number of 1\'s in every row and column, store them in a separate array and later whenever we encounter a 1, look up the value in its column and row (our two arrays) , if both of them are <=1, increment the count.\n\n```\n public int numSpecial(int[][] mat... | 5 | 0 | [] | 0 |
special-positions-in-a-binary-matrix | ✅ One Line Solution | one-line-solution-by-mikposp-gp1k | (Disclaimer: these are not examples to follow in a real project - they are written for fun and training mostly. PEP 8 is violated intentionally)\n\n# Code #1\nT | MikPosp | NORMAL | 2023-12-13T09:49:05.475037+00:00 | 2024-01-13T10:45:35.449724+00:00 | 767 | false | (Disclaimer: these are not examples to follow in a real project - they are written for fun and training mostly. PEP 8 is violated intentionally)\n\n# Code #1\nTime complexity: $$O(m*n)$$. Space complexity: $$O(1)$$.\n```\nclass Solution:\n def numSpecial(self, M: List[List[int]]) -> int:\n return sum(sum(rr[j... | 4 | 0 | ['Array', 'Matrix', 'Python', 'Python3'] | 1 |
special-positions-in-a-binary-matrix | C++ solutions | c-solutions-by-infox_92-wahb | \nclass Solution {\npublic:\n\tint numSpecial(vector<vector<int>>& mat) {\n\t\tint nRows = mat.size();\n\t\tint nCols = mat[0].size();\n\t\tint res = 0;\n\t\tfo | Infox_92 | NORMAL | 2022-11-09T06:41:33.670990+00:00 | 2022-11-09T06:41:33.671033+00:00 | 717 | false | ```\nclass Solution {\npublic:\n\tint numSpecial(vector<vector<int>>& mat) {\n\t\tint nRows = mat.size();\n\t\tint nCols = mat[0].size();\n\t\tint res = 0;\n\t\tfor(int i = 0; i < nRows; i++){\n\t\t\tfor(int j = 0; j < nCols; j++){\n\t\t\t\tif(mat[i][j] == 1){\n\t\t\t\t\tint colSum = 0;\n\t\t\t\t\tint rowSum = 0;\n\t\t... | 4 | 0 | ['C', 'C++'] | 0 |
special-positions-in-a-binary-matrix | 📌 Python Memory-Efficient Solution | python-memory-efficient-solution-by-vand-v79k | \nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n onesx = []\n onesy = []\n for ri, rv in enumerate(mat):\n | vanderbilt | NORMAL | 2022-02-27T04:18:16.261610+00:00 | 2022-03-01T09:34:25.617743+00:00 | 629 | false | ```\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n onesx = []\n onesy = []\n for ri, rv in enumerate(mat):\n for ci, cv in enumerate(rv):\n if cv == 1:\n onesx.append(ri)\n onesy.append(ci)\n \n ... | 4 | 0 | ['Python', 'Python3'] | 1 |
special-positions-in-a-binary-matrix | Simple Java Solution O(m*n) (Easy) | simple-java-solution-omn-easy-by-shantan-xkvc | Up-vote if helpfull\n\nclass Solution {\n public int numSpecial(int[][] mat) {\n int row[]=new int[mat.length];\n int col[]=new int[mat[0].leng | ShantanuSinghStrive | NORMAL | 2021-05-25T10:48:04.522140+00:00 | 2021-05-25T10:48:04.522181+00:00 | 533 | false | **Up-vote if helpfull**\n```\nclass Solution {\n public int numSpecial(int[][] mat) {\n int row[]=new int[mat.length];\n int col[]=new int[mat[0].length];\n int res=0;\n for(int i=0;i<mat.length;i++){\n for(int j=0;j<mat[0].length;j++){\n if(mat[i][j]==1){\n ... | 4 | 1 | ['Java'] | 0 |
special-positions-in-a-binary-matrix | [Python3] 5-line 2-pass | python3-5-line-2-pass-by-ye15-iwag | \nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n m, n = len(mat), len(mat[0])\n row, col = [0]*m, [0]*n\n for i, | ye15 | NORMAL | 2020-09-13T04:01:48.682002+00:00 | 2020-09-13T04:01:48.682049+00:00 | 672 | false | ```\nclass Solution:\n def numSpecial(self, mat: List[List[int]]) -> int:\n m, n = len(mat), len(mat[0])\n row, col = [0]*m, [0]*n\n for i, j in product(range(m), range(n)): \n if mat[i][j]: row[i], col[j] = 1 + row[i], 1 + col[j]\n return sum(mat[i][j] and row[i] == 1 and col[... | 4 | 1 | ['Python3'] | 0 |
special-positions-in-a-binary-matrix | JavaScript | C++ | javascript-c-by-nurliaidin-15k1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Nurliaidin | NORMAL | 2023-12-14T13:53:32.066143+00:00 | 2023-12-14T13:53:32.066182+00:00 | 293 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['C++', 'JavaScript'] | 2 |
special-positions-in-a-binary-matrix | Beats 100% of users with C++ ( BruteForce Approach ) Easy to understand | beats-100-of-users-with-c-bruteforce-app-hdo2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n\n\n\n\nif you like the approach please upvote it\n\n\n\n\n\n\nif you like the appr | abhirajpratapsingh | NORMAL | 2023-12-13T16:19:55.131442+00:00 | 2023-12-13T16:19:55.131473+00:00 | 16 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n\n\n\n\nif you like the approach please upvote it\n\n\n\n\n\n\nif you like the approach please upvote it\n\n\n\n\n\n# Ap... | 3 | 0 | ['Array', 'Matrix', 'C++'] | 0 |
special-positions-in-a-binary-matrix | Java Approach||beginners friendly | java-approachbeginners-friendly-by-hithe-dpxs | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Hitheash | NORMAL | 2023-12-13T14:57:48.835962+00:00 | 2023-12-13T14:57:48.835986+00:00 | 651 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Java'] | 0 |
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